Integration by partial fraction problem (∫dx/x(x^2 + 4)^2)

[Bimpo]

Homework Statement

I came across a problem that I can't solve
and it is ∫dx/x(x^2 + 4)^2

Homework Equations

None

The Attempt at a Solution

So I'm pretty sure this is to be solved by partial fraction since I am on a chapter on
Integration by partial fraction.

so I started with A/x + (Bx+C/x^2+4) + [Dx+E/(x^2 +4)^2]

and I get a reeeeaallllyy loooonnngg equation once I go around that
Am I on the right track? Or did I make a mistake? is this even to be solved by partial fraction?

 

[Vorde]

First rewrite as $\int$$\frac{1}{x(x^2 + 4)^2}$ dx

Then turn into partial fractions (lets ignore the integral part for now and focus on the fraction): $\frac{A}{x}$ + $\frac{B}{(x^2 + 4)^2}$ = $\frac{1}{x(x^2 + 4)^2}$

A $(x^2 + 4)^2$ + B$x$ = 1

Solve for A and B and plug back in.

 

[SammyS]

Homework Statement

I came across a problem that I can't solve
and it is ∫dx/x(x^2 + 4)^2

Homework Equations

None

The Attempt at a Solution

So I'm pretty sure this is to be solved by partial fraction since I am on a chapter on
Integration by partial fraction.

so I started with A/x + (Bx+C/x^2+4) + [Dx+E/(x^2 +4)^2]

and I get a reeeeaallllyy loooonnngg equation once I go around that
Am I on the right track? Or did I make a mistake? is this even to be solved by partial fraction?

First try the substitution u = x² .

You will still get to work with partial fractions, but they won't be quite as complicated.

I assume your problem is actually $\displaystyle \int\ \frac{dx}{x(x^2+4)^2}\,.$

Parentheses are important.

 

[vela]

so I started with A/x + (Bx+C/x^2+4) + [Dx+E/(x^2 +4)^2]

and I get a reeeeaallllyy loooonnngg equation once I go around that
Am I on the right track? Or did I make a mistake? is this even to be solved by partial fraction?

You should have written A/x + (Bx+C)/(x^2+4) + (Dx+E)/(x^2+4)^2. As SammyS said, parentheses are important.

Your expansion is fine. If you stick with this approach, you should find A=1/16, B=-1/16, C=0, D=-1/4, and E=0.

 

[Bimpo]

sorry for late response but thanks for the replies